The demand function for a certain item is x = (p + 3)e^-(p+0.3) in thousands of

Question

The demand function for a certain item is x = (p + 3)e^-(p+0.3) in thousands of items demanded at a price of p dollars per item. Use interval notation to indicate the range of prices corresponding to elastic, inelastic, and unitary price elasticity of demand. You use 'i' for infinity and '-i for -infinity. And use 'U' for the union symbol. The range of prices so that demand is elastic is The range of prices so that demand is inelastic is a) The unit price necessary for unitary elasticity is You have attempted this problem 0 times. You have unlimited attempts remaining.

Explanation / Answer

Differentiate the given function to get...

dx/dp = e^(-(p + 0.3)) + (p + 3) * -e^(-(p + 0.3))

Now, the price of elasticity is...

y(p) = (e^(-(p + 0.3)) + (p + 3) * -e^(-(p + 0.3))) * p/((p + 3)e^(-(p + 0.3))

To answer the parts of the question, we must know that...

d = 0 - perfectly inelastic

|d| < 1 - inelastic

d = 1 - unit elastic

|d| > 1 - elastic

d = infinity [perfectly elastic]

Elastic: | e^ - (p+0.3) + (p+3)*(-1)e^-(p+0.3)*p/(p+3)e^-(p+0.3)|>1

             = | {p (e^- p-0.3 - e^-p-0.3 (p+3)} / (p+3) | e^Re(p) + 0.3 > 1

             = | p+3| < | p (p+2) |

             = | {p(p+2)} / (p+3) | >1

             = 1.34986 | {p( e^ - p- 0.3 - e^ -p-0.3 ( p+3) } / (p+3) | e^Re (p) >1

              when p is positive

   p > 1.30278

   Re (p) > 1/2 ( sqrt 13 - 1)

        -3 < Re (p) < 1/2 ( -1 - sqrt 13)

         Re(p) < -3

        Re(p) = -3 Im (p) > 0

         Re(p) = -3 Im (p) < 0

hence interval notation will be:

     ( 1/2 ( sqrt 13 - 1) , infinity ) , ( - infinity , -3)

Inelastic:   | e^ - (p+0.3) + (p+3)*(-1)e^-(p+0.3)*p/(p+3)e^-(p+0.3)| < 1

             = | {p (e^- p-0.3 - e^-p-0.3 (p+3)} / (p+3) | e^Re(p) + 0.3 < 1

           = | p+3| > | p (p+2) |

             = | {p(p+2)} / (p+3) | <1

             = 1.34986 | {p( e^ - p- 0.3 - e^ -p-0.3 ( p+3) } / (p+3) | e^Re (p) < 1

             =    when p is positive

   p < 1.30278

- 2.30278 < p < 1.30278

- 2.30278 < Re (p) < 1.30278

- sqrt { 0.5( -2 Re(p)^2 - 4 Re(p) - 3) + 0.5 sqrt (16 Re(p)^2 + 48 Re(p) +45)} < Im (p) < sqrt [ (0.5( -2Re(p)^2 - 4 Re (p) -3) + 0.5 sqrt ( 16 Re(p)^2 + 48 (p) + 45 ) ]

Then integers are : p = +-1 , -2 , 0

hence interval notation will be:

     ( 2.30278 , 1.30278 )

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